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$Box A$: 2 gray balls 1 white ball $Box B$: 1 gray ball 3 white balls. We choose randomly a box and choose randomly one ball. (1) Calculate the probability that the ball is white. My attempt:$ 0.5*(1/3)+0.5*(3/4)=13/24$ (2) Calculate the probability that Box A was chosen when it is known that white ball was chosen. My attempt: $[0.5*(1/3)]/(first question)$ (3) We return the ball to the box from which it was taken and now we take randomly another ball . calculate the probability that the other ball is white when it is known that the first one is white. My attempt: $(first question)*(first question)/(first question)=13/24 $(4) Like in the third question except now the first ball stays out. My attempt: ${0.5*(1/3)+0.5*0.5+0.5*(3/4)}/(first question)$

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The answers to a) and b) are correct. We look at c).

Given that the chosen ball was white, the probability it came from A is $p_A=\frac{(0.5)(1/3)}{13/24}$, and the probability it came from B is $p_B=\frac{(0.5)(3/4)}{13/24}$.

Thus the (conditional) probability the next ball is white is $(1/3)p_A+(3/4)p_B$.

For d), the $1/3$ is replaced by $0/2$ and the $3/4$ is replaced by $2/3$.

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  • $\begingroup$ @ André Nicolas about d) Doesn't it miss the case when the first was chosen from A but the second from B? $\endgroup$ – user188750 Nov 16 '14 at 7:43
  • $\begingroup$ I interpreted "take another ball" in both c) and d) as meaning that we take another ball from the box from which we drew in the first place. But the other interpretation is possible, the phrasing is a bit ambiguous. $\endgroup$ – André Nicolas Nov 16 '14 at 7:49

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